Chapter 31, Problem 52E

(a)

To determine

Amount of radium required to produce as many decays per second.

Answer to Problem 52E

An amount of $1.011\times {10}^7\text{kg}$ radium is required to produce as many decays per second.

Explanation of Solution

Nuclear reactor has accumulated radioactive waste of ${10}^{10}\text{Ci}$.

Write the expression of activity $a$ of radium.

  $a=\frac{\left(0.693\right)mN}{t}$

Here, $t$ is half-life of radium, $m$ is number of moles of radium and $N$ is Avogadro’s number.

Rearrange the above expression

  $m=\frac{at}{\left(0.693\right)N}$

Since each mole of radium has mass $0.226\text{kg}$ therefore mass of the radium $M$ is:

  $M=\left(0.226\text{kg}\cdot {\text{mole}}^{-1}\right)m$

Substitute $\frac{at}{\left(0.693\right)N}$ for $m$ in above expression

  $M=\left(0.226\text{kg}\cdot {\text{mole}}^{-1}\right)\left(\frac{at}{\left(0.693\right)N}\right)$        (I)

Conclusion:

The value of source activity $a$ is:

  $\begin{array}{c}a=\left({10}^{10}\text{Ci}\left(\frac{3.70\times {10}^{10}{\text{s}}^{-1}}{1\text{Ci}}\right)\right)\\ =3.70\times {10}^{20}{\text{s}}^{-1}\end{array}$

The value of half-life $t$ is:

  $\begin{array}{c}t=\left(1.60\times {10}^3\text{year}\left(\frac{3.15\times {10}^7\text{s}}{1\text{year}}\right)\right)\\ =5.04\times {10}^{10}\text{s}\end{array}$

Substitute $5.04\times {10}^{10}\text{s}$ for $t$, $3.70\times {10}^{20}{\text{s}}^{-1}$ for $a$ and $6.02\times {10}^{23}{\text{mole}}^{-1}$ for $N$ in expression (I)

  $\begin{array}{c}M=\left(0.226\frac{\text{kg}}{\text{mole}}\right)\left(\frac{\left(\frac{3.70\times {10}^{20}}{\text{s}}\right)\left(5.04\times {10}^{10}\text{s}\right)}{\left(0.693\right)\left(\frac{6.02\times {10}^{23}}{\text{mole}}\right)}\right)\\ =1.011\times {10}^7\text{kg}\end{array}$

Thus, $1.011\times {10}^7\text{kg}$ of radium is required to produce as many decays per second.

(b)

To determine

Rate of production of heat by reactor.

Answer to Problem 52E

The rate of production of heat by reactoris $59.2\text{MW}$.

Explanation of Solution

An energy of $1\text{MeV}$ is assumed to be released by one decay.

Write the expression of rate of energy produced by reactor

  $E=ae$        (II)

Here, $e$ is energy released per decay.

Conclusion:

Substitute $1\text{MeV}$ for $e$ and $3.70\times {10}^{20}{\text{s}}^{-1}$ for $a$ in expression (II)

  $\begin{array}{l}E=\left(3.70\times {10}^{20}{\text{s}}^{-1}\right)\left(1\text{MeV}\left(\frac{1.6\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\left(\frac{1\text{W}}{1\text{J}\cdot {\text{s}}^{-1}}\right)\left(\frac{1\text{MW}}{{10}^6\text{W}}\right)\right)\\ =59.2\text{MW}\end{array}$

Thus, reactor produces heat at a rate $59.2\text{MW}$.

(c)

To determine

Ratio of this power to rate of $3\times {10}^9\text{watts}$.

Answer to Problem 52E

The desired power ratio is $0.0197$.

Explanation of Solution

Write the expression of ratio

  $r=\frac{E}{\left(3\times {10}^9\text{W}\right)}$        (III)

Here, $r$ is the desired ratio.

Conclusion:

Substitute $59.2\text{MW}$ for $E$ in expression (III)

  $\begin{array}{c}r=\frac{\left(59.2\text{MW}\left(\frac{{10}^6\text{W}}{1\text{MW}}\right)\right)}{\left(3\times {10}^9\text{W}\right)}\\ =0.0197\end{array}$

Thus, the desired power ratio is $0.0197$.